Question 6 of 6: Transmitter Power Budget for DSB, and Square-Wave AM/DSB Envelopes

Question 6: Transmitter Power Budget for DSB, and Square-Wave AM/DSB Envelopes (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Transmitter limits $S_T\le3$ kW (average power), $A_{max}^2\le8$ kW (peak envelope power); tone message with amplitude $A_m=1$; part (c) uses a $\pm1$ square-wave message instead of a tone.

Find. (a) $S_x$; (b) the maximum achievable power per sideband $P_{sb}$ for DSB under both limits simultaneously; (c) sketches of $x_c(t)$, envelopes dashed, for square-wave AM ($\mu=0.5$, $\mu=1$) and square-wave DSB.

Approach. (a)–(b): write $S_T$ and $A_{max}$ for tone-modulated DSB in terms of the carrier amplitude $A_c$, find which of the two given limits binds, and use that to get the maximum $P_{sb}$. (c): for a $\pm1$ square wave the AM/DSB envelope is piecewise constant, switching instantly between two levels every half period.

1. Part (a) — message power. For a tone $x(t)=A_m\cos(\omega_mt)$ the average power is $S_x=\dfrac{A_m^2}{2}$; with $A_m=1$, $$S_x=\boxed{0.5}\ \ (\text{normalized units}).$$
2. Part (b) — maximum sideband power. A DSB signal $x_c(t)=A_c\cos(\omega_mt)\cos(\omega_ct)=\tfrac{A_c}{2}\cos[(\omega_c-\omega_m)t]+\tfrac{A_c}{2}\cos[(\omega_c+\omega_m)t]$ has envelope $A_c|\cos\omega_mt|$, so the peak envelope power is $A_{max}^2=A_c^2$ (since $A_m=1$), and the average transmitted power is $S_T=\tfrac{A_c^2}{2}S_x=\tfrac{A_c^2}{4}$. Each sideband has amplitude $A_c/2$, so its power is $P_{sb}=(A_c/2)^2/2=A_c^2/8=S_T/2$ (the two sidebands split the total power equally — DSB carries no separate discrete carrier term). Checking which constraint binds: $A_c^2\le8$ kW from the peak-power limit gives $S_T\le2$ kW, already under the $3$ kW average-power ceiling, so the peak-power limit is the tighter one ($A_c^2\le12$ kW would be needed to hit $S_T=3$ kW, more than the $8$ kW allowed). Using the binding $A_c^2=8$ kW, $$P_{sb}=\frac{A_c^2}{8}=\frac{8\ \text{kW}}{8}=\boxed{1\ \text{kW}},\qquad S_T=\frac{A_c^2}{4}=\boxed{2\ \text{kW}}\ (\le3\ \text{kW, not binding}).$$
3. Part (c) — square-wave envelope sketches. With $x(t)=\pm1$ switching every half period, the AM envelope $A_c[1+\mu x(t)]$ takes only two values: for $\mu=0.5$ the envelope alternates between $1.5A_c$ and $0.5A_c$ (both positive — safe, under-100% modulation); for $\mu=1$ it alternates between $2A_c$ and exactly $0$ (the critical, 100%-modulation case, where the carrier is fully suppressed on every other half-cycle). For DSB, $x_c(t)=A_c\,x(t)\cos(\omega_ct)$ has constant envelope magnitude $A_c$ throughout — what switches every half period is not the amplitude but the carrier's phase, which flips by $180^{\circ}$ each time $x(t)$ changes sign. All three cases are sketched below, dashed lines marking the envelope.